find all values of x for which the perimeter is at most 32
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Answer:
0 < x ≤ 11
Step-by-step explanation:
perimeter = (x - 6) + (x - 6) + x + x
⇒ perimeter = 4x - 12
If the perimeter is "at most 32" then
4x - 12 ≤ 32
⇒ 4x ≤ 44
⇒ x ≤ 11
Therefore, 0 < x ≤ 11